cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A237719 Numbers n such that k(n) = (n(n+1)/2 mod n) = (antisigma(n) mod n) + (sigma(n) mod n).

Original entry on oeis.org

1, 2, 6, 12, 18, 20, 24, 28, 30, 40, 42, 54, 56, 66, 70, 78, 80, 88, 100, 102, 104, 112, 114, 120, 126, 138, 140, 150, 160, 162, 174, 176, 180, 186, 196, 198, 200, 204, 208, 220, 222, 224, 228, 234, 240, 246, 258, 260, 272, 276, 282, 294, 304, 306, 308, 318, 320
Offset: 1

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Author

Jaroslav Krizek, Mar 16 2014

Keywords

Comments

Numbers n such that k(n) = A142150(n) = A229110(n) + A054024(n).
Numbers n such that k(n) = (A000217(n) mod n) = (A024816(n) mod n) + (A000203(n) mod n).
k(n) = 0 for odd n, k(n) = n/2 for even n.
If there are any odd multiply-perfect numbers, they are members of this sequence.
If there is no odd multiply-perfect number, then:
(1) the only odd number in this sequence is 1,
(2) corresponding sequence of numbers k(n): {0; a(n) / 2 for n > 1}.
Supersequence of A159907, A007691 and A000396.

Examples

			12 is in the sequence because k(12) = (12*(12+1)/2) mod 12 = antisigma(12) mod 12 + sigma(12) mod 12; k(12) = 6 = 4 + 2 = n/2.

Crossrefs

Cf. A000203, A000217, A024816, A054024, A142150, A229110.

Programs

  • Magma
    [n: n in [1..320] | IsZero(n*(n+1)div 2 mod n - SumOfDivisors(n) mod n - (n*(n+1)div 2-SumOfDivisors(n)) mod n)]

A295236 Hemi-imperfect numbers: numbers such that the denominator of k/A206369(k) is equal to 2.

Original entry on oeis.org

3, 10, 42, 60, 63, 840, 1260, 12642, 18480, 18900, 18963, 154350, 228480, 252840, 379260, 3458700, 5562480, 5688900, 68772480, 1041068700, 15032156160, 53621568000, 4524679004160, 9812746944000
Offset: 1

Views

Author

Michel Marcus, Nov 19 2017

Keywords

Comments

This is to rho (A206369) what hemiperfect numbers are to sigma (A000203).
After 3, 10 and 42, whose quotients are resp. 3/2, 5/2 and 7/2, 373316437260251755241798182764378479569038727298776522806597255168000000 is an instance of a term with quotient 9/2. - Michel Marcus, Dec 17 2017
a(25) > 10^13. - Giovanni Resta, Feb 17 2020

Examples

			3 is a term since rho(3) = 2, so 3/rho(3) is 3/2.
10 is a term since rho(10) = 4, so 10/rho(10) is 5/2.
42 is a term since rho(42) = 12, so 42/rho(42) is 7/2.

Links

Crossrefs

Cf. A127724 (k-imperfect), A206369 (rho).
Cf. A159907 (hemiperfect).

Programs

  • Maple
    rho:= proc(n) local f;
  mul((f[1]^(f[2]+1)+(-1)^f[2])/(f[1]+1), f = ifactors(n)[2]);
end proc:
select(t -> denom(t/rho(t)) = 2, [$1..10^6]); # Robert Israel, Nov 20 2017
  • Mathematica
    (* b = A209369 *) b[n_] := n*DivisorSum[n, LiouvilleLambda[#]/# &];
Select[Range[10^6], If[Denominator[#/b[#]] == 2, Print[#]; True, False]&] (* Jean-François Alcover, Dec 04 2017 *)
  • PARI
    rho(n) = {my(f = factor(n), res = q = 1); for(i=1, #f~, q = 1; for(j = 1, f[i, 2], q = -q + f[i, 1]^j); res *= q); res;}
isok(n) = denominator(n/rho(n))==2;

Extensions

a(20) from Jinyuan Wang, Feb 15 2020
a(21)-a(24) from Giovanni Resta, Feb 17 2020

A218429 Numbers k for which sigma(k)/k - 7/8 is an integer.

Original entry on oeis.org

8, 760320, 1468800, 4612608, 1414886400, 4935598080, 83655936000, 172888934400, 173172916224000, 225464073246720, 575926998958080, 650264059920384, 73439222840111923200, 88160928190086850560, 450645911113324953600, 600860926790121553920
Offset: 1

Views

Author

Zdenek Cervenka, Oct 28 2012

Keywords

Comments

a(8) > 10^11. - Donovan Johnson, Nov 01 2012
a(9) > 10^12. - Giovanni Resta, Nov 04 2012
Note that there are no terms here with abundancy 23/8 (k=2). - Michel Marcus, Jun 26 2013

Crossrefs

Cf. A218404, A159907, A218405.

Extensions

a(5)-a(7) from Donovan Johnson, Nov 01 2012
a(8) from Giovanni Resta, Nov 04 2012
More terms from Michel Marcus, Jun 26 2013

A227882 Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2.

Original entry on oeis.org

1, 3, 19, 0, 87, 117, 0, 30, 0, 0
Offset: 2

Views

Author

Michel Marcus, Oct 25 2013

Keywords

Comments

The hemiperfect that are obtained are coprime to p = 2*n-1.
When p=2*n-1 is prime, if m is a n-multiperfect is such that valuation(m, p) = 1, then let's define k = m/p, sigma(k) = sigma(m/p) = sigma(m)/sigma(p) = (n*m)/(p+1) = (n*m)/(2*n) = m/2. So sigma(k)/k = m/(2*k) = (k*p)/(2*k) = p/2 = (2*n-1)/2.

Examples

			a(2) = 1, since the only perfect number multiple of 3 is 6, and 6/3=2 has abundancy 3/2.
a(3) = 3, since the 3 known hemiperfect of abundancy 5/2 are coprime to 5.
a(5) = a(8) = a(11) = 0, since for those n, 2*n-1 is not prime.
a(10) is also 0, since all known 10-multiperfect are at least divisible by 19^2.

Links

Crossrefs

Cf. A000396 (2), A005820 (3), A027687 (4), A046060 (5), A046061 (6), A007691 (integer abundancy).
Cf. A141643 (5/2), A055153 (7/2), A141645 (9/2), A159271 (11/2), A160678 (13/2), A159907 (half-integer abundancy).
Cf. A006254.
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